A mixture of 2.3 g formic acid (HCOOH) and 4.5 g oxalic acid ((COOH)2) is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. The weight (in g) of the remaining gas is

Correct answer: (A) 2.8. Answer: (A) 2.8 g Solution: Conc. H2SO4 dehydrates both acids: HCOOH ->[H2SO4] CO + H2O and (COOH)2 ->[H2SO4] CO + CO2 + H2O Moles of HCOOH = 2.346 = 0.05 mol 0.05 mol CO. Moles of (COOH)2 = 4.590 = 0.05 mol 0.05 mol CO + 0.05 mol CO2. KOH absorbs only CO2. Remaining CO = 0.10 mol = 0.10 × 28 = 2.8 g.

NEET 2018 · ChemistryPrevious Year Question

A mixture of 2.3 g formic acid ( $HCOOH$ ) and 4.5 g oxalic acid ( $(COOH) X_{2}$ ) is treated with conc. $H X_{2} SO X_{4}$ . The evolved gaseous mixture is passed through $KOH$ pellets. The weight (in g) of the remaining gas is

Answer: (A) 2.8. \textbf{Answer:} (A) 2.8 g \textbf{Solution:} Conc.

A.2.8✓
B.3.0
C.1.4
D.4.4

Correct Answer

(A) 2.8

Solution & Explanation

\textbf{Answer:} (A) 2.8 g \textbf{Solution:} Conc. $H X_{2} SO X_{4}$ dehydrates both acids: $HCOOH H X_{2} SO X_{4} CO + H X_{2} O$ and $(COOH) X_{2} H X_{2} SO X_{4} CO + CO X_{2} + H X_{2} O$ Moles of $HCOOH$ $= \frac{2.3}{46} = 0.05$ mol $\Rightarrow$ 0.05 mol $CO$ . Moles of $(COOH) X_{2}$ $= \frac{4.5}{90} = 0.05$ mol $\Rightarrow$ 0.05 mol $CO$ + 0.05 mol $CO X_{2}$ . $KOH$ absorbs only $CO X_{2}$ . Remaining $CO$ $= 0.10$ mol $= 0.10 \times 28 = 2.8$ g.

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A mixture of 2.3 g formic acid (HCOOH) and 4.5 g oxalic acid ((COOH)X2​) is treated with conc. HX2​SOX4​. The evolved gaseous mixture is passed through KOH pellets. The weight (in g) of the remaining gas is

Solution & Explanation

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A mixture of 2.3 g formic acid ( $HCOOH$ ) and 4.5 g oxalic acid ( $(COOH) X_{2}$ ) is treated with conc. $H X_{2} SO X_{4}$ . The evolved gaseous mixture is passed through $KOH$ pellets. The weight (in g) of the remaining gas is